imone/ARB huggingface dataset:
Question: Compute$$ \int_{0}^{\pi} \frac{x \sin x}{1+\sin ^{2} x} d x . $$
Answer: We use the example from the introduction for the particular function $f(x)=\frac{x}{1+x^{2}}$ to transform the integral into$$ \pi \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\sin ^{2} x} d x . $$This is the same as$$ \pi \int_{0}^{\frac{\pi}{2}}-\frac{d(\cos x)}{2-\cos ^{2} x}, $$which with the substitution $t=\cos x$ becomes$$ \pi \int_{0}^{1} \frac{1}{2-t^{2}} d t=\left.\frac{\pi}{2 \sqrt{2}} \ln \frac{\sqrt{2}+t}{\sqrt{2}-t}\right|_{0} ^{1}=\frac{\pi}{2 \sqrt{2}} \ln \frac{\sqrt{2}+1}{\sqrt{2}-1} . $$
Question: Compute up to two decimal places the number$$ \sqrt{1+2 \sqrt{1+2 \sqrt{1+\cdots+2 \sqrt{1+2 \sqrt{1969}}}}} $$where the expression contains 1969 square roots.
Answer: Let$$ x_{n}=\sqrt{1+2 \sqrt{1+2 \sqrt{1+\cdots+2 \sqrt{1+2 \sqrt{1969}}}}} $$with the expression containing $n$ square root signs. Note that$$ x_{1}-(1+\sqrt{2})=\sqrt{1969}-(1+\sqrt{2})<50 . $$Also, since $\sqrt{1+2(1+\sqrt{2})}=1+\sqrt{2}$, we have $$ \begin{aligned} x_{n+1}-(1+\sqrt{2})=& \sqrt{1+2 x_{n}}-\sqrt{1+2(1+\sqrt{2})}=\frac{2\left(x_{n}-(1-\sqrt{2})\right)}{\sqrt{1+2 x_{n}}+\sqrt{1+2(1+\sqrt{2})}} \\ &<\frac{x_{n}-(1+\sqrt{2})}{1+\sqrt{2}} . \end{aligned} $$From here we deduce that$$ x_{1969}-(1+\sqrt{2})<\frac{50}{(1+\sqrt{2})^{1968}}<10^{-3}, $$and the approximation of $x_{1969}$ with two decimal places coincides with that of $1+\sqrt{2}=$ 2.41. This argument proves also that the limit of the sequence is $1+\sqrt{2}$.
Question: Let $M_{2 \times 2}$ be the vector space of all real $2 \times 2$ matrices. Let

$$
A=\left(\begin{array}{cc}
1 & 2 \\
-1 & 3
\end{array}\right) \quad B=\left(\begin{array}{cc}
2 & 1 \\
0 & 4
\end{array}\right)
$$

and define a linear transformation $L: M_{2 \times 2} \rightarrow M_{2 \times 2}$ by $L(X)=A X B$. Compute the trace of $L$.


Answer:  Identify $M_{2 \times 2}$ with $\mathbb{R}^{4}$ via

$$
\left(\begin{array}{ll}
a & b \\nc & d
\end{array}\right) \leftrightarrow\left(\begin{array}{l}
a \\nb \\nc \\nd
\end{array}\right)
$$

and decompose $L$ into the multiplication of two linear transformations,

$$
M_{2 \times 2} \simeq \mathbb{R}^{4} \stackrel{L_{A}}{\longrightarrow} \mathbb{R}^{4} \stackrel{L_{B}}{\longrightarrow} \mathbb{R}^{4} \simeq M_{2 \times 2}
$$

where $L_{A}(X)=A X$ and $L_{B}(X)=X B$. is

The matrices of these two linear transformations on the canonical basis of $\mathbb{R}^{4}$

$$
L_{A}=\left(\begin{array}{rrrr}
1 & 0 & 2 & 0 \\n0 & 1 & 0 & 2 \\n-1 & 0 & 3 & 0 \\n0 & -1 & 0 & 3
\end{array}\right) \quad \text { and } L_{B}=\left(\begin{array}{llll}
2 & 0 & 0 & 0 \\n1 & 4 & 0 & 0 \\n0 & 0 & 2 & 0 \\n0 & 0 & 1 & 4
\end{array}\right)
$$

then $\operatorname{det} L=\operatorname{det} L_{A} \cdot \operatorname{det} L_{B}=(9+6+2(2+3)) \cdot(2 \cdot 32)=2^{6} \cdot 5^{2}$, and to compute the trace of $L$, we only need the diagonal elements of $L_{A} \cdot L_{B}$, that is,

$$
\operatorname{tr} L=2+4+6+12=24 \text {. }
$$
Question: Define

$$
F(x)=\int_{\sin x}^{\cos x} e^{\left(t^{2}+x t\right)} d t .
$$

Compute $F^{\prime}(0)$.


Answer:  Let

$$
G(u, v, x)=\int_{v}^{u} e^{t^{2}+x t} d t .
$$

Then $F(x)=G(\cos x, \sin x, x)$, so

$$
\begin{aligned}
F^{\prime}(x) &=\frac{\partial G}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial G}{\partial v} \frac{\partial v}{\partial x}+\frac{\partial G}{\partial x} \\n&=e^{u^{2}+x u}(-\sin x)-e^{\left(v^{2}+x v\right)} \cos x+\int_{v}^{u} t e^{t^{2}+x t} d t
\end{aligned}
$$

and

$$
F^{\prime}(0)=-1+\int_{0}^{1} t e^{t^{2}} d t=\frac{1}{2}(e-3) .
$$
Question: What is the probability that the sum of two randomly chosen numbers in the interval $[0,1]$ does not exceed 1 and their product does not exceed $\frac{2}{9}$?
Answer: Let $x$ and $y$ be the two numbers. The set of all possible outcomes is the unit square$$ D=\{(x, y) \mid 0 \leq x \leq 1,0 \leq y \leq 1\} . $$The favorable cases consist of the region$$ D_{f}=\left\{(x, y) \in D \mid x+y \leq 1, x y \leq \frac{2}{9}\right\} . $$This is the set of points that lie below both the line $f(x)=1-x$ and the hyperbola $g(x)=\frac{2}{9 x}$. equal toThe required probability is $P=\frac{\operatorname{Area}\left(D_{f}\right)}{\operatorname{Area}(D)}$. The area of $D$ is 1 . The area of $D_{f}$ is$$ \int_{0}^{1} \min (f(x), g(x)) d x . $$The line and the hyperbola intersect at the points $\left(\frac{1}{3}, \frac{2}{3}\right)$ and $\left(\frac{2}{3}, \frac{1}{3}\right)$. Therefore,$$ \operatorname{Area}\left(D_{f}\right)=\int_{0}^{1 / 3}(1-x) d x+\int_{1 / 3}^{2 / 3} \frac{2}{9 x} d x+\int_{2 / 3}^{1}(1-x) d x=\frac{1}{3}+\frac{2}{9} \ln 2 . $$We conclude that $P=\frac{1}{3}+\frac{2}{9} \ln 2 \approx 0.487$.
Question: Evaluate

$$
\int_{0}^{2 \pi} e^{\left(e^{i \theta}-i \theta\right)} d \theta .
$$


Answer:  By Cauchy's Integral Formula for derivatives, we have

therefore,

$$
\left.\frac{d}{d z} e^{z}\right|_{z=0}=\frac{1}{2 \pi i} \int_{|z|=1} \frac{e^{z}}{z^{2}} d z=\frac{1}{2 \pi} \int_{0}^{2 \pi} e^{e^{i \theta}-i \theta} d \theta
$$

$$
\int_{0}^{2 \pi} e^{e^{i \theta}-i \theta} d \theta=2 \pi .
$$
Question: Compute the product$$ \left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right) \cdots $$
Answer: For $N \geq 2$, define$$ a_{N}=\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right) \cdots\left(1-\frac{4}{(2 N-1)^{2}}\right) . $$The problem asks us to find $\lim _{N \rightarrow \infty} a_{N}$. The defining product for $a_{N}$ telescopes as follows:$$ \begin{aligned} a_{N} &=\left[\left(1-\frac{2}{1}\right)\left(1+\frac{2}{1}\right)\right]\left[\left(1-\frac{2}{3}\right)\left(1+\frac{2}{3}\right)\right] \cdots\left[\left(1-\frac{2}{2 N-1}\right)\left(1+\frac{2}{2 N-1}\right)\right] \\ &=(-1 \cdot 3)\left(\frac{1}{3} \cdot \frac{5}{3}\right)\left(\frac{3}{5} \cdot \frac{7}{5}\right) \cdots\left(\frac{2 N-3}{2 N-1} \cdot \frac{2 N+1}{2 N-1}\right)=-\frac{2 N+1}{2 N-1} . \end{aligned} $$Hence the infinite product is equal to$$ \lim _{N \rightarrow \infty} a_{N}=-\lim _{N \rightarrow \infty} \frac{2 N+1}{2 N-1}=-1 . $$
Question: Suppose the coefficients of the power series

$$
\sum_{n=0}^{\infty} a_{n} z^{n}
$$

are given by the recurrence relation

$$
a_{0}=1, a_{1}=-1,3 a_{n}+4 a_{n-1}-a_{n-2}=0, n=2,3, \ldots .
$$

Find the radius of convergence $r$.


Answer:  From the recurrence relation, we see that the coefficients $a_{n}$ grow, at most, at an exponential rate, so the series has a positive radius of convergence. Let $f$ be the function it represents in its disc of convergence, and consider the polynomial $p(z)=3+4 z-z^{2}$. We have

$$
\begin{aligned}
p(z) f(z) &=\left(3+4 z-z^{2}\right) \sum_{n=0}^{\infty} a_{n} z^{n} \\n&=3 a_{0}+\left(3 a_{1}+4 a_{0}\right) z+\sum_{n=0}^{\infty}\left(3 a_{n}+4 a_{n-1}-a_{n-2}\right) z^{n} \\n&=3+z
\end{aligned}
$$

So

$$
f(z)=\frac{3+z}{3+4 z-z^{2}} .
$$

The radius of convergence of the series is the distance from 0 to the closest singularity of $f$, which is the closest root of $p$. The roots of $p$ are $2 \pm \sqrt{7}$. Hence, the radius of convergence is $\sqrt{7}-2$.
Question: For integers $n \geq 2$ and $0 \leq k \leq n-2$, compute the determinant$$ \left|\begin{array}{ccccc} 1^{k} & 2^{k} & 3^{k} & \cdots & n^{k} \\ 2^{k} & 3^{k} & 4^{k} & \cdots & (n+1)^{k} \\ 3^{k} & 4^{k} & 5^{k} & \cdots & (n+2)^{k} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n^{k} & (n+1)^{k} & (n+2)^{k} & \cdots & (2 n-1)^{k} \end{array}\right| . $$
Answer: The polynomials $P_{j}(x)=(x+j)^{k}, j=0,1, \ldots, n-1$, lie in the $(k+1)$-dimensional real vector space of polynomials of degree at most $k$. Because $k+1<n$, they are linearly dependent. The columns consist of the evaluations of these polynomials at $1,2, \ldots, n$, so the columns are linearly dependent. It follows that the determinant is zero.
Question: Compute$$ \lim _{n \rightarrow \infty}\left[\frac{1}{\sqrt{4 n^{2}-1^{2}}}+\frac{1}{\sqrt{4 n^{2}-2^{2}}}+\cdots+\frac{1}{\sqrt{4 n^{2}-n^{2}}}\right] . $$
Answer: We have$$ \begin{aligned} s_{n} &=\frac{1}{\sqrt{4 n^{2}-1^{2}}}+\frac{1}{\sqrt{4 n^{2}-2^{2}}}+\cdots+\frac{1}{\sqrt{4 n^{2}-n^{2}}} \\ &=\frac{1}{n}\left[\frac{1}{\sqrt{4-\left(\frac{1}{n}\right)^{2}}}+\frac{1}{\sqrt{4-\left(\frac{2}{n}\right)^{2}}}+\cdots+\frac{1}{\sqrt{4-\left(\frac{n}{n}\right)^{2}}}\right] . \end{aligned} $$Hence $s_{n}$ is the Riemann sum of the function $f:[0,1] \rightarrow \mathbb{R}, f(x)=\frac{1}{\sqrt{4-x^{2}}}$ associated to the subdivision $x_{0}=0<x_{1}=\frac{1}{n}<x_{2}=\frac{2}{n}<\cdots<x_{n}=\frac{n}{n}=1$, with the intermediate points $\xi_{i}=\frac{i}{n} \in\left[x_{i}, x_{i+1}\right]$. The answer to the problem is therefore$$ \lim _{n \rightarrow \infty} s_{n}=\int_{0}^{1} \frac{1}{\sqrt{4-x^{2}}} d x=\left.\arcsin \frac{x}{2}\right|_{0} ^{1}=\frac{\pi}{6} . $$
Question: Consider the sequences $\left(a_{n}\right)_{n}$ and $\left(b_{n}\right)_{n}$ defined by $a_{1}=3, b_{1}=100, a_{n+1}=3^{a_{n}}$, $b_{n+1}=100^{b_{n}}$. Find the smallest number $m$ for which $b_{m}>a_{100}$. 
Answer: We need to determine $m$ such that $b_{m}>a_{n}>b_{m-1}$. It seems that the difficult part is to prove an inequality of the form $a_{n}>b_{m}$, which reduces to $3^{a_{n-1}}>100^{b_{m-1}}$, or $a_{n-1}>\left(\log _{3} 100\right) b_{m-1}$. Iterating, we obtain $3^{a_{n-2}}>\left(\log _{3} 100\right) 100^{b_{m-2}}$, that is,$$ a_{n-2}>\log _{3}\left(\log _{3} 100\right)+\left(\left(\log _{3} 100\right) b_{m-2} .\right. $$Seeing this we might suspect that an inequality of the form $a_{n}>u+v b_{n}$, holding for all $n$ with some fixed $u$ and $v$, might be useful in the solution. From such an inequality we would derive $a_{n+1}=3^{a_{n}}>3^{u}\left(3^{v}\right)^{b_{m}}$. If $3^{v}>100$, then $a_{n+1}>3^{u} b_{m+1}$, and if $3^{u}>u+v$, then we would obtain $a_{n+1}>u+v b_{m+1}$, the same inequality as the one we started with, but with $m+1$ and $n+1$ instead of $m$ and $n$.The inequality $3^{v}>100$ holds for $v=5$, and $3^{u}>u+5$ holds for $u=2$. Thus $a_{n}>2+5 b_{m}$ implies $a_{n+1}>2+5 b_{m+1}$. We have $b_{1}=100, a_{1}=3, a_{2}=27, a_{3}=3^{27}$, and $2+5 b_{1}=502<729=3^{6}$, so $a_{3}>2+5 b_{1}$. We find that $a_{n}>2+5 b_{n-2}$ for all $n \geq 3$. In particular, $a_{n} \geq b_{n-2}$.On the other hand, $a_{n}<b_{m}$ implies $a_{n+1}=3^{a_{n}}<100^{b_{m}}<b_{m+1}$, which combined with $a_{2}<b_{1}$ yields $a_{n}<b_{n-1}$ for all $n \geq 2$. Hence $b_{n-2}<a_{n}<b_{n-1}$, which implies that $m=n-1$, and for $n=100, m=99$.
Question: Let the sequence $a_{0}, a_{1}, \ldots$ be defined by the equation

$$
1-x^{2}+x^{4}-x^{6}+\cdots=\sum_{n=0}^{\infty} a_{n}(x-3)^{n} \quad(0<x<1) .
$$

Find

$$
\limsup _{n \rightarrow \infty}\left(\left|a_{n}\right|^{\frac{1}{n}}\right) .
$$


Answer:  As

$$
1-x^{2}+x^{4}-x^{6}+\cdots=\frac{1}{1+x^{2}}
$$

which has singularities at $\pm i$, the radius of convergence of

$$
\sum_{n=0}^{\infty} a_{n}(x-3)^{n}
$$

is the distance from $3$ to $\pm i, and |3 \mp i|=\sqrt{10}$. We then have

$$
\limsup _{n \rightarrow \infty}\left(\left|a_{n}\right|^{\frac{1}{n}}\right)=\frac{1}{\sqrt{10}} \cdot
$$
Question: Find the radius of convergence $R$ of the Taylor series about $z=1$ of the function $f(z)=1 /\left(1+z^{2}+z^{4}+z^{6}+z^{8}+z^{10}\right)$. 


Answer:  The rational function

$$
f(z)=\frac{1-z^{2}}{1-z^{12}}
$$

has poles at all nonreal twelfth roots of unity (the singularities at $z^{2}=1$ are removable). Thus, the radius of convergence is the distance from 1 to the nearest singularity:

$$
R=|\exp (\pi i / 6)-1|=\sqrt{(\cos (\pi / 6)-1)^{2}+\sin ^{2}(\pi / 6)}=\sqrt{2-\sqrt{3}} .
$$
Question: Find the number of permutations $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ of the numbers $1,2,3,4,5,6$ that can be transformed into $1,2,3,4,5,6$ through exactly four transpositions (and not fewer).
Answer: The condition from the statement implies that any such permutation has exactly two disjoint cycles, say $\left(a_{i_{1}}, \ldots, a_{i_{r}}\right)$ and $\left(a_{i_{r+1}}, \ldots, a_{i_{6}}\right)$. This follows from the fact that in order to transform a cycle of length $r$ into the identity $r-1$, transpositions are needed. Moreover, we can only have $r=5,4$, or 3 .When $r=5$, there are $\left(\begin{array}{l}6 \\ 1\end{array}\right)$ choices for the number that stays unpermuted. There are $(5-1) !$ possible cycles, so in this case we have $6 \times 4 !=144$ possibilities. When $r=4$, there are $\left(\begin{array}{l}6 \\ 4\end{array}\right)$ ways to split the numbers into the two cycles (two cycles are needed and not just one). One cycle is a transposition. There are $(4-1) !=6$ choices for the other. Hence in this case the number is 90 . Note that here exactly four transpositions are needed.Finally, when $r=3$, then there are $\left(\begin{array}{l}6 \\ 3\end{array}\right) \times(3-1) ! \times(3-1) !=40$ cases. Therefore, the answer to the problem is $144+90+40=274$.
Question: Given the fact that $\int_{-\infty}^{\infty} e^{-x^{2}} d x=\sqrt{\pi}$, evaluate the integral

$$
I=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^{2}+(y-x)^{2}+y^{2}\right)} d x d y .
$$
Answer:  We write

$$
\begin{aligned}
I &=\int_{-\infty}^{\infty} e^{-3 y^{2} / 2}\left(\int_{-\infty}^{\infty} e^{-2 x^{2}+2 x y-y^{2} / 2} d x\right) d y \\n&=\int_{-\infty}^{\infty} e^{-3 y^{2} / 2}\left(\int_{-\infty}^{\infty} e^{-2\left(x-\frac{y}{2}\right)^{2}} d x\right) d y
\end{aligned}
$$

making the substitution $t=\sqrt{2}\left(x-\frac{y}{2}\right), d t=\sqrt{2} d x$ on the inner integral, we get

$$
\begin{aligned}
I &=\frac{1}{\sqrt{2}} \int_{-\infty}^{\infty} e^{-3 y^{2} / 2}\left(\int_{-\infty}^{\infty} e^{-t^{2}} d t\right) d y \\n&=\sqrt{\frac{\pi}{2}} \int_{-\infty}^{\infty} e^{-3 y^{2} / 2} d y
\end{aligned}
$$

now making the substitution: $s=\sqrt{\frac{3}{2}} y, d s=\sqrt{\frac{3}{2}} d y$ we obtain

$$
\begin{aligned}
I &=\sqrt{\frac{\pi}{3}} \int_{-\infty}^{\infty} e^{-s^{2}} d s \\n&=\frac{\pi}{\sqrt{3}} .
\end{aligned}
$$
Question: The zeros of the polynomial $P(x)=x^{3}-10 x+11$ are $u$, $v$, and $w$. Determine the value of $\arctan u+\arctan v+\arctan w$.
Answer: First solution: Let $\alpha=\arctan u, \beta=\arctan v$, and $\arctan w$. We are required to determine the sum $\alpha+\beta+\gamma$. The addition formula for the tangent of three angles, $$ \tan (\alpha+\beta+\gamma)=\frac{\tan \alpha+\tan \beta+\tan \gamma-\tan \alpha \tan \beta \tan \gamma}{1-(\tan \alpha \tan \beta+\tan \beta \tan \gamma+\tan \alpha \tan \gamma)}, $$implies$$ \tan (\alpha+\beta+\gamma)=\frac{u+v+w-u v w}{1-(u v+v w+u v)} . $$Using Viète's relations,$$ u+v+w=0, \quad u v+v w+u w=-10, \quad u v w=-11, $$we further transform this into $\tan (\alpha+\beta+\gamma)=\frac{11}{1+10}=1$. Therefore, $\alpha+\beta+\gamma=\frac{\pi}{4}+k \pi$, where $k$ is an integer that remains to be determined.From Viète's relations we can see the product of the zeros of the polynomial is negative, so the number of negative zeros is odd. And since the sum of the zeros is 0 , two of them are positive and one is negative. Therefore, one of $\alpha, \beta, \gamma$ lies in the interval $\left(-\frac{\pi}{2}, 0\right)$ and two of them lie in $\left(0, \frac{\pi}{2}\right)$. Hence $k$ must be equal to 0 , and $\arctan u+$ $\arctan v+\arctan w=\frac{\pi}{4}$.Second solution: Because$$ \operatorname{Im} \ln (1+i x)=\arctan x, $$we see that$$ \begin{aligned} \arctan u+\arctan v+\arctan w &=\operatorname{Im} \ln (i P(i))=\operatorname{Im} \ln (11+11 i) \\ &=\arctan 1=\frac{\pi}{4} . \end{aligned} $$
Question: Find the maximum of $x_{1}^{3}+\ldots+x_{10}^{3}$ for $x_{1}, \ldots, x_{10} \in[-1,2]$ such that  $$ x_{1}+\ldots+x_{10}=10.$$
Answer: Look at the behavior of expression $x_{1}^{3}+x_{2}^{3}$ when we move the variables $x_{1} \leq x_{2}$ together, that is, replace them with $x_{1}+\varepsilon$ and $x_{2}-\varepsilon, 0<\varepsilon \leq \frac{x_{2}-x_{1}}{2}$, and when we move them apart, that is, replace them with $x_{1}-\varepsilon$ and $x_{2}+\varepsilon, \varepsilon>0$. After moving together, the product $x_{1} x_{2}$ is not decreasing, and after moving apart, it is not increasing. Hence the sum of cubes of two variables   $$ x_{1}^{3}+x_{2}^{3}=\left(x_{1}+x_{2}\right)\left(\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right) $$  is not decreasing as two variables move apart, if $x_{1}+x_{2} \geq 0$, and as they move together, if $x_{1}+x_{2} \leq 0$.  We will move apart couples of variables $x_{i}, x_{j}$, for which $x_{i}+x_{j} \geq 0$, until one of the variables coincides with an end of the segment $[-1 ; 2]$. That can be done while at least one couple of variables with nonnegative sum remains in the interval $(-1 ; 2)$. After that we get one of the next two situations:  (1) some variables are equal to 2 or $-1$, and the rest of the variables have negative pairwise sums (and moreover each of the variables in the interval $(-1 ; 2)$ has negative sum with -1), or  (2) some variables are equal to 2 or $-1$, and a single variable in the interval $(-1 ; 2)$ has nonnegative sum with $-1$.  In the first case, let $x$ be the mean value of all the variables in $[-1 ; 2)$. Then $x<0$ because all pairwise sums are negative. We will move together couples of variables from the interval $[-1 ; 2)$ as described above, until one of the variables in the couple equals $x$.  Thus, we can start with an arbitrary collection of points $x_{1}, \ldots, x_{10}$ and use the described above moving together and apart (which preserve the sum of elements and not decrease the sum of cubes) to reach one of the following collections:  (1) $k$ variables are at a point $x \in[-1 ; 0]$ and $10-k$ variables are at the point 2 $(k=0, \ldots, 10)$  (2) $k$ variables are at the point $-1$, a single variable equals $x \in[1 ; 2]$, and $9-k$ variables are at the point $2(k=0, \ldots, 9)$.  From the conditions $x_{1}+\ldots+x_{10}=10$ and $x \in[-1 ; 0]$ (or $x \in[1 ; 2]$ ), we obtain that either $k=4$ or $k=5$ for collections of the first type and $k=3$ for collections of the second type. It remains to examine the following collections:  $$ \begin{gathered} \left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, 2,2,2,2,2,2\right), \quad(0,0,0,0,0,2,2,2,2,2), \\ (-1,-1,-1,1,2,2,2,2,2,2) . \end{gathered} $$  The maximal value of the sum of cubes is equal to $47.5$ and attained at the first collection.  Answer: $47
Question: Two airplanes are supposed to park at the same gate of a concourse. The arrival times of the airplanes are independent and randomly distributed throughout the 24 hours of the day. What is the probability that both can park at the gate, provided that the first to arrive will stay for a period of two hours, while the second can wait behind it for a period of one hour?
Answer: The set of possible events is modeled by the square $[0, 24] \times[0,24]$. It is, however, better to identify the 0th and the 24th hours, thus obtaining a square with opposite sides identified, an object that in mathematics is called a torus (which is, in fact, the Cartesian product of two circles. The favorable region is outside a band of fixed thickness along the curve $x=y$ on the torus as depicted in Figure 110. On the square model this region is obtained by removing the points $(x, y)$ with $|x-y| \leq 1$ together with those for which $|x-y-1| \leq 1$ and $|x-y+1| \leq 1$. The required probability is the ratio of the area of the favorable region to the area of the square, and is$$ P=\frac{24^{2}-2 \cdot 24}{24^{2}}=\frac{11}{12} \approx 0.917 . $$
Question: Find a limit  $$ \lim _{n \rightarrow \infty}\left(\int_{0}^{1} e^{x^{2} / n} d x\right)^{n}. $$
Answer: Fix an arbitrary $\varepsilon>0$. Since the function $y=e^{t}$ is convex, for $t \in[0, \varepsilon]$ its graph lies above the tangent $y=1+t$ but under the secant $y=1+a_{\varepsilon} t$, where $a_{\varepsilon}=\frac{1}{\varepsilon}\left(e^{\varepsilon}-1\right)$. Hence for $\frac{1}{n} \leq \varepsilon$ it holds  $$ \left(\int_{0}^{1} e^{x^{2} / n} d x\right)^{n} \geq\left(\int_{0}^{1}\left(1+\frac{x^{2}}{n}\right) d x\right)^{n}=\left(1+\frac{1}{3 n}\right)^{n} \rightarrow e^{1 / 3}, \text { as } n \rightarrow \infty, $$  and  $$ \left(\int_{0}^{1} e^{x^{2} / n} d x\right)^{n} \leq\left(\int_{0}^{1}\left(1+\frac{a_{\varepsilon} x^{2}}{n}\right) d x\right)^{n}=\left(1+\frac{a_{\varepsilon}}{3 n}\right)^{n} \rightarrow e^{a_{\varepsilon} / 3}, \text { as } n \rightarrow \infty $$  Since $a_{\varepsilon}=\frac{1}{\varepsilon}\left(e^{\varepsilon}-1\right) \rightarrow 1$, as $\varepsilon \rightarrow 0+$, we get $\lim _{n \rightarrow \infty}\left(\int_{0}^{1} e^{x^{2} / n} d x\right)^{n}=e^{1 / 3}$.  Answer: $e^{1 / 3}$
Question: Let $x_{0}=1$ and

$$
x_{n+1}=\frac{3+2 x_{n}}{3+x_{n}}, \quad n \geqslant 0 .
$$

Find the limit $x_{\infty}=\lim _{n \rightarrow \infty} x_{n}$.


Answer:  Obviously, $x_{n} \geqslant 1$ for all $n$; so, if the limit exists, it is $\geqslant 1$, and we can pass to the limit in the recurrence relation to get

$$
x_{\infty}=\frac{3+2 x_{\infty}}{3+x_{\infty}} \text {; }
$$

in other words, $x_{\infty}^{2}+x_{\infty}-3=0$. So $x_{\infty}$ is the positive solution of this quadratic equation, that is, $x_{\infty}=\frac{1}{2}(-1+\sqrt{13})$.

To prove that the limit exists, we use the recurrence relation to get

$$
\begin{aligned}
x_{n+1}-x_{n} &=\frac{3+2 x_{n}}{3+x_{n}}-\frac{3+2 x_{n-1}}{3+x_{n-1}} \\n&=\frac{3\left(x_{n}-x_{n-1}\right)}{\left(3+x_{n}\right)\left(3+x_{n+1}\right)}
\end{aligned}
$$

Hence, $\left|x_{n+1}-x_{n}\right| \leqslant \frac{1}{3}\left|x_{n}-x_{n-1}\right|$. Iteration gives

$$
\left|x_{n+1}-x_{n}\right| \leqslant 3^{-n}\left|x_{1}-x_{0}\right|=\frac{1}{3^{n} \cdot 4} .
$$

The series $\sum_{n=1}^{\infty}\left(x_{n+1}-x_{n}\right)$, of positive terms, is dominated by the convergent series $\frac{1}{4} \sum_{n=1}^{\infty} 3^{-n}$ and so converges. We have $\sum_{n=1}^{\infty}\left(x_{n+1}-x_{n}\right)=\lim x_{n}-x_{1}$ and we are done.
Question: Consider a sequence $x_{n}=x_{n-1}-x_{n-1}^{2}, n \geq 2, x_{1} \in(0,1)$. Calculate  $$ \lim _{n \rightarrow \infty} \frac{n^{2} x_{n}-n}{\ln n} . $$
Answer: By the monotone convergence theorem, it is easy to show that $x_{n} \rightarrow 0$, as $n \rightarrow \infty$. Next, by the Stolz-Cesaro theorem it holds  $$ \begin{aligned} n x_{n}=\frac{n}{\frac{1}{x_{n}}} \sim \frac{1}{\frac{1}{x_{n}}-\frac{1}{x_{n-1}}}=\frac{x_{n-1} x_{n}}{x_{n-1}-x_{n}}=\\ \quad=\frac{x_{n-1}\left(x_{n-1}-x_{n-1}^{2}\right)}{x_{n-1}^{2}}=1-x_{n-1} \rightarrow 1, \text { as } n \rightarrow \infty . \end{aligned} $$  Transform the expression in question and use the Stolz-Cesaro theorem once again:   $$ \begin{aligned} \frac{n^{2} x_{n}-n}{\ln n}=\frac{n x_{n} \cdot\left(n-\frac{1}{x_{n}}\right)}{\ln n} \sim \frac{n-\frac{1}{x_{n}}}{\ln n} \sim \frac{1-\frac{1}{x_{n}}+\frac{1}{x_{n-1}}}{\ln n-\ln (n-1)}=\frac{1-\frac{1}{x_{n}}+\frac{1}{x_{n-1}}}{-\ln \left(1-\frac{1}{n}\right)} \sim \\ \sim n\left(1-\frac{1}{x_{n-1}-x_{n-1}^{2}}+\frac{1}{x_{n-1}}\right)=-\frac{n x_{n-1}}{1-x_{n-1}} \rightarrow-1, \text { as } n \rightarrow \infty \end{aligned} $$  Answer: $-1